If $X\sim N(\mu,\sigma^2)$ and $Y=e^X$, then what is the median of $Y$?

I am pretty sure that $Y$ is also distributed normal. To try to prove it, I attempted both the method of moment generating functions and the method of cdfs. I just can't get it. Thanks for your help. Once I show that, getting the median is the same as getting the mean.

  • $\begingroup$ "I am pretty sure that Y is also distributed normal." Ouch! That hurts... Note that Y is always nonnegative. $\endgroup$ – Did Sep 12 '14 at 19:48
  • $\begingroup$ Dag nabbit, this is why I should never graph functions of pdfs in my TI-84 $\endgroup$ – nonremovable Sep 12 '14 at 19:51
  • $\begingroup$ If half the mass of $X$ is below $\mu$, then half the mass of $e^X$ is below $e^\mu$. $\endgroup$ – André Nicolas Sep 12 '14 at 20:29
  • $\begingroup$ Some comments in this thread are now deleted, which originated from the OP's comment that "this is why (they) should never graph functions of pdfs in my TI-84". These were underlining the intrinsic uselessness of graph calculators to approach a whole class of mathematical questions, to which the present one belongs. As such, I cannot fathom why one saw fit to delete these comments (especially in view of how fiercely the OP resisted to the point explained to them and how widespread the misconceptions the exchange revealed are). $\endgroup$ – Did Sep 13 '14 at 13:33

Hint: let $\Phi$ be the CDF of $Z \sim \mathcal{N}\left(0, 1\right)$. Notice that $$\mathbb{P}\left(Y \leq y\right) = \mathbb{P}\left(e^{X} \leq y\right) = \mathbb{P}\left(X \leq \ln(y)\right) = \mathbb{P}\left(Z \leq \dfrac{\ln(y)-\mu}{\sigma}\right)= \Phi\left[\dfrac{\ln(y)-\mu}{\sigma}\right]\text{.}$$

  • $\begingroup$ Thanks for that. I got as far as that but then I didn't know how to continue. $\endgroup$ – nonremovable Sep 12 '14 at 19:59
  • $\begingroup$ Suppose $y$ is the median. What would the above equation have to be equal to? $\endgroup$ – Clarinetist Sep 12 '14 at 19:59
  • $\begingroup$ $\phi\left[\frac{\ln(\mu)-\mu}{\sigma}\right]$ $\endgroup$ – nonremovable Sep 12 '14 at 20:02
  • $\begingroup$ Nope. What is the definition of median? $\endgroup$ – Clarinetist Sep 12 '14 at 20:02
  • $\begingroup$ wait no, because $Z$ is standard. $\endgroup$ – nonremovable Sep 12 '14 at 20:03


  • What is a median of $X$ normal $N(\mu,\sigma^2)$?
  • If $m$ is a median of the random variable $U$, and $V=A(U)$ where the function $A:\mathbb R\to\mathbb R$ is nondecreasing, what would be a median of $V$?
  • $\begingroup$ The median is the value at which probabilities are 1/2 total on each side. For your second question, I don't know. I was tempted to say it should be the same since the pdf of $X$ is symmetrical about the $y$ axis. $\endgroup$ – nonremovable Sep 12 '14 at 19:50
  • $\begingroup$ This answers none of the two points. $\endgroup$ – Did Sep 12 '14 at 19:51
  • $\begingroup$ You just changed your two points! The answer to your new first point is $\mu$, again I don't know about the second point. $\endgroup$ – nonremovable Sep 12 '14 at 19:53
  • $\begingroup$ Not changed, paraphrased, to force you to read them. $\endgroup$ – Did Sep 12 '14 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.